Question on SWR
 

Ian White GM3SEK wrote in
:

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

I think the ARRL graph is based on a well known, but apparently not well
understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

On terminology, I prefer to not use the term 'extra loss due to VSWR'
because the name implies to many, that it is always positive. IMHO a
better way to speak of the loss is as line loss under mismatched
conditions... and those conditions are more specific than just a VSWR
figure.

In a lot of cases, the approximation is sufficiently accurate... but you
lose visibility of the error when you assume that the approximation is
ALWAYS sufficiently accurate, a Rule of Thumb or ROT for short.

Owen

Question on SWR
 

Owen Duffy wrote in
:

....
I think the ARRL graph is based on a well known, but apparently not
well understood formula.

The only text book that I can recall spelling out the assumptions that
underly the integral that produces the formula is Philip Smith's 'The
Electronic Applications of the Smith Chart'.

The formula is developed by integrating I^2 over an electrical half wave
of line on one side or the other from the observation point, according
to PS. (I don't know the origin of the formula, I am not suggesting that
PS invented it, not that he didn't.)

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave. A requirement for this is that line loss
must be relatively low, that the exponential real term in the
transmission line equations is close to zero.

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

For a line that is many half waves, the formula is fine so long as VSWR
is approximately constant (now a very low loss line). If the line is
longer than many half waves, but not an exact integral number of half
waves, then the error in the partial section will be somewhat diminished
relatively by the loss in the complete half wave sections.

If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

So, in summary, the formula is good for low loss half wave lines, or
even longish random length low loss lines, but not good for short random
length lines or very long lines.

So, why is the formula so popular?

Could it be that it underpins one of the popular myths of ham radio,
that VSWR necessarily increases line loss?

Modern computation tools are better than the 70 year old graphical
method. Publication of the formula without qualification with the
underlying assumptions treats the reader as a dummy.

Owen

Question on SWR
 

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

Question on SWR
 

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the current
and voltage maxima and minima and the relative contribution of R and G
to loss. So, the formula may have significant error for short lines that
are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same at
both ends for it to not matter which end is the observation point, so
therefore the first assumption is that VSWR is approximately equal at
both ends of the half wave.
If a practical line is very long, it cannot qualify as having a constant
VSWR (unless it is 1, in which case the formula is unnecessary), so the
formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at the antenna,
and that it DOES take into account that, for a lossy line, the line portions
closer to the transmitter are subjected to a lower SWR.

I try to explain my argument. Let us assume that the line consists of the
cascade of many identical line pieces, each having a 1-dB loss, that one can
freely add or remove. Adding a piece causes an increase of line loss by 1dB +
some extra loss due to SWR. If you add many 1-dB pieces (that corresponds to
increasing the total line loss), the chart shows that the extra loss caused by
the last added pieces gets smaller and smaller (the chart curves all tend to
saturate for an increasing line loss), and this could be explained by the fact
that the ARRL formula does take into account the fact that the last added pieces
are subjected to a lower SWR (and hence yield a lower extra loss) .

What do you think about that?

73

Tony I0JX

Question on SWR
 

"Antonio Vernucci" wrote in
:

I full agree with your statement:

If the line section is not exactly a half wave, then the real loss
factor might be higher or lower depending on the location of the
current and voltage maxima and minima and the relative contribution
of R and G to loss. So, the formula may have significant error for
short lines that are not exactly a half wave.

But I am not certain about this other statement:

Straight away, that tells you that the VSWR must be almost the same
at both ends for it to not matter which end is the observation point,
so therefore the first assumption is that VSWR is approximately equal
at both ends of the half wave.
If a practical line is very long, it cannot qualify as having a
constant VSWR (unless it is 1, in which case the formula is
unnecessary), so the formula is not suited.

I have a feeling that the ARRL chart makes reference to the SWR at
the antenna, and that it DOES take into account that, for a lossy
line, the line portions closer to the transmitter are subjected to a
lower SWR.

Approximations that depend on VSWR as a load metric can:
1. depend on the integral over a half wave of very low loss line then
apply it as a constant loss loss per unit length; or
2. treat the forward and reflected waves as waves independently subject
to a constant loss per unit length.

I explained the sources of error in extending (1) to the general case in
my earlier post.

Case (2) assumes that the attenuation is the result of (vector) addition
of the power that is lost independently from each travelling wave at any
point, whereas the power lost is due to the effect of currents and
voltage resulting from vector addition of the voltages and currents of
the two waves at each point.

In some scenarios, they may be good approximations, but there are also
scenarios where they are poor approximations. (I gave an example in an
earlier posting where they both fail.)

The reality is that on a practical mismatched line, loss per unit length
is not constant with displacement. See my notes on VSWR at varying
displacement on a practical line, see
http://www.vk1od.net/VSWR/displacement.htm .

Look at Fig 9.

Note that the loss vs displacement line does not have a constant slope,
and anything that ignores that is ignoring an aspect of the problem.

Note that in the example, the red line dips below the blue line (meaning
loss under mismatched conditions is LESS than matched line loss at some
lengths). Any method that prevents that result is ignoring an aspect of
the problem.

The loss under mismatch conditions does depend on load impedance, and if
you throw away some of the detail and reduce it to load VSWR, then you
increase the scope for error.

Is there application for the approximations? Certainly, I use them... but
in the knowledge that they are approximations and an awareness of where
they are not good approximations and may not produce an adequate answer
for the current problem.

Your original posting was about reconciling the chart with some examples,
and I noted that the chart itself is a source of significant error in
some scenarios.

BTW, your calculations seem to fall into case (2), and if so, are subject
to the same errors... though they may reconcile well with a chart based
on that approximation.

Owen